Problem:
A person has a net worth of $0$. In their first month they increase their net worth be $100$. In the second month they increase their net worth by $200$. In the third month they increase their net worth by $300$. Hence after three months their net worth will be $600$.
What will their net worth be after $n$ months?
Answer:
Let $N$ be their net worth after $n$ months. \begin{align*} A &= \sum_{i=1}^{n} \sum_{j=1}^{i} 100 \\ A &= \sum_{i=1}^{n} 50(i)(i+1) \\ A &= 50 \sum_{i=1}^{n} i^2 + i \\ A &= 50 \left( \dfrac{ n(n+1)(2n+1) }{6} + \dfrac{n(n+1)}{2} \right) \\ A &= 50 \left( \dfrac{ n(n+1)(2n+1) }{6} + \dfrac{3n^2+3n)}{6} \right) \\ A &= \left( \dfrac{25}{3}\right) ( (n^2+n)(2n+1) + 3n^2 + 3n ) \\ A &= \left( \dfrac{25}{3}\right) ( 2n^3 + 2n^2 + n^2 + n + 3n^2 + 3n ) \\ A &= \left( \dfrac{25}{3}\right) ( 2n^3 + 6n^2 + 4n ) \\ \end{align*} Hence the answer is: $$ A = \left( \dfrac{50}{3}\right) ( n^3 + 3n^2 + 2n ) $$
As a partial check, I will evaluate $A$ when $n = 2$. \begin{align*} A &= \ \left( \dfrac{50}{3}\right) ( 2^3 + 3(2^2) + 2(2) ) \\ A &= \ \left( \dfrac{50}{3}\right) ( 8 + 12 + 4 ) \\ A &= 400 \end{align*} However, I should have gotten $300$.
As a partial check, I will evaluate $A$ when $n = 4$. \begin{align*} A &= \left( \dfrac{50}{3}\right) ( (4)^3 + 3(4)^2 + 2(4) ) \\ A &= \left( \dfrac{50}{3}\right) ( 64 + 3(16)+ 8 ) \\ A &= \left( \dfrac{50}{3}\right) ( 120 ) = 50(40) = 2000 \end{align*} However, I should have gotten $1000$. Where did I go wrong?
Here is an updated answer, which I now believe is right.
Let $N$ be his net worth after $n$ months. \begin{align*} A &= \sum_{i=1}^{n} 100i \\ A &= \dfrac{ 100(n)(n+1)}{2} \\ A &= 50n^2 + 50n \end{align*} Hence the answer is: $$ A = 50n^2 + 50n $$
As a partial check, I will evaluate $A$ when $n = 2$. \begin{align*} A &= 50(2)^2 + 50(2) = 50(4) + 100 \\ A &= 300 \end{align*} It checks.
As a partial check, I will evaluate $A$ when $n = 4$. \begin{align*} A &= 50(4)^2 + 50(4) \\ A &= 50(16) + 200 = 1000 \end{align*} It checks.
Observe that $A$ increases as $n$ increase. It does not increase exponential but it does increase faster than linearly. That is what I would expect in the answer.
Now let's look at some special cases. The first is when $n = 0$. In this case, we have: $$ A = 50(0^2) + 50(0) = 0 $$ Hence in this case, the formula checks. Now let's consider the case where $n = 6$. The right answer is: $$100+200+300+400+500+600 = 2100$$ We have: \begin{align*} A &= 50n^2 + 50n \\ A &= 50(6^2) + 50(6) = 50(36) + 300 \\ A &= 2100 \\ \end{align*} Hence it checks. Therefore, I believe the second solution posted is correct.